From the foregoing explanation of the manner of determining the
trigonometrical functions of angles, the methods of solving right-angled
triangles will be readily perceived, and only a few examples need
therefore be given.
Let _a_ and _b_ represent the sides and _c_ the hypothenuse of a
right-angled triangle, and _a_° and _b_° the angles opposite to the
sides. Then of the possible cases we will take
(1.) Given _c_ and _a_°, to find _a_, _b_, and _b_°.
The angle _b_° = 90 − _a_°, while _a_ = _c_ sin _a_° and _b_ = _c_ sin
_b_°. To find _a_, therefore, the index of S is set to _c_ on A, and the
value of _a_ read on A opposite _a_° on S. In the same manner the value
of _b_ is obtained.
EX.—Given in a right-angled triangle _c_ = 9 ft. and _a_° = 30°. Find
_a_, _b_, and _b_°.
The angle _b_° = 90 − 30 = 60°. To find _a_, set R.H. index of S to 9
on A, and over 30° on S read _a_ = 4·5 ft. on A. Also, with the slide
in the same position, read _b_ = 7·8 ft. [7·794] on A over 60° on S.
(2.) Given _a_ and _c_, to determine _a_°, _b_°, and _b_.
In this case advantage is taken of the fact that in every triangle the
sides are proportional to the sines of the opposite angles. Therefore,
as in this case the hypothenuse c subtends a right angle, of which the
sine = 1, the R.H. index (or 90°) on S is set to the length of _c_ on A,
when under _a_ on A is found _a_° on S. Hence _b_° and _b_ may be
determined.
(3.) Given _a_ and _a_°, to find _b_, _c_, and _b_°.
Here _b_° = (90 − _a_°), and the solution is similar to the foregoing.
(4.) Given _a_ and _b_, to find _a_°, _b_°, and _c_.
To find _a_°, we have tan. _a_° = _a_/_b_, which in the above example
will be (4·5)/(7·8) = 0·577. Therefore, placing the slide so that the
indices of T coincide with those of D, we read opposite 0·577 on D the
value of _a_° = 30°. The hypothenuse _c_ is readily obtained from _c_ =
_a_/(sin _a_°).